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Unformatted text preview: Rate Laws
• Rate laws are always determined experimentally unless you have been
told a ratelimiting or slow step. • Reaction order is always defined in terms of reactant (not product)
concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the
reactant in the balanced chemical equation.
ln[A]t =
kt + ln[A]0 1/[A]t = kt + 1/[A]0 [A]t =
kt + [A]0 Figure 16.5 Integrated rate laws and reac=on orders. ln[A]t =
kt + ln[A]0 1/[A]t = kt + 1/[A]0 [A]t =
kt + [A]0 Figure 16.6 Graphical determina=on of the reac=on order for the decomposi=on of N2O5. Figure 16.7 A plot of [N2O5] vs. =me for three half
lives. for a ﬁrst
order process t1/2 = ln 2 k = 0.693 k • Half
life is deﬁned as the =me required for one
half of a reactant to react. • Because [A] at t1/2 is one
half of the original [A], [A]t = 0.5 [A]0. Sample Problem 16.6 PROBLEM: Using Molecular Scenes to Determine Half
Life Compound A (red) converts to compound B (black) in a ﬁrst
order process. The following scenes represent the reac=on mixture at the start of the reac=on (at 0.0 s) and aaer 30.0 s: (a) Find the half
life, t1/2, of the reac=on. (b) Calculate the rate constant, k. (c) Draw a scene that represents the reac=on mixture aaer 2.00 min. PLAN: (a) Half
life is constant for ﬁrst
order reac=on. The =me it takes for half of the red spheres to become black is the half
life. (b) We subs=tute the value of t1/2 from part (a) into Equa=on 16.7 and solve for k. (c) Knowing the half
life, one can calculate how many red and black spheres would be in the mixture at 2.00 min. Sample Problem 16.6 Using Molecular Scenes to Determine Half
Life con=nued SOLUTION: (a) (b) At t = 0.0 s, there are 8 A (red) molecules and no B (black). At t = 30.0 s, there are 6 A and 2 B molecules present, so ¼ of A has reacted. Therefore, t1/2 = (2 x 30.0 s) = 60.0 s For ﬁrst
order reac=on, t1/2 = 0.693/k k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10
2 s
1 (c) The =me given, t = 120. s, is two half
lives. There should be ½(1/2) molecules of A (red) lea. Sample Problem 16.7 PROBLEM: Determining the Half
Life of a First
Order Reac=on Cyclopropane is the smallest cyclic hydrocarbon. Because its 60o bond angles reduce orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000oC via the following ﬁrst
order reac=on: The rate constant is 9.2 s
1, (a) What is the half
life of the reac=on? (b) How long does it take for the concentra=on of cyclopropane to reach one
quarter of the ini=al value? 0.693 PLAN: Use the half
life equa=on, t1/2 = , to ﬁnd the half
life. k One
quarter of the ini=al value means two half
lives have passed. SOLUTION: (a) 0.693
t1/2 =
1 = 0.075 s 9.2 s (b) 2 t1/2 = 2(0.075 s) = 0.15 s Half
Life
2nd order For a second
order process, set [A]t=0.5 [A]0 in 2nd order equa=on. Figure 16.8 Dependence of the rate constant on temperature. The Eﬀect of Temperature on Reac=on Rate The Arrhenius Equa=on where k is the kine=c rate constant at T Ea is the ac=va=on energy R is the universal gas constant ln k = ln A
Ea/RT ln k2 k1 =
Ea R T is the Kelvin temperature A is the collision frequency factor 1
T2 1 T1 Figure 16.9 Graphical determina=on of the ac=va=on energy. ln k =
Ea/R (1/T) + ln A Sample Problem 16.8 PROBLEM: Determining the Energy of Ac=va=on The decomposi=on of hydrogen iodide, 2HI(g) H2(g) + I2(g) has rate constants of 9.51 x 10
9 L/mol•s at 500. K and 1.10 x 10
5 L/mol•s at 600. K. Find Ea. PLAN: Use the modiﬁca=on of the Arrhenius equa=on to ﬁnd Ea. SOLUTION: ln k2 k1 = – Ea = –(8.314 J/mol K) Ea 1 R T2 ln – 1
T1 1.10x10
5 L/mol s 9.51x10
9 L/mol s Ea = 1.76 x 105 J/mol = 176 kJ/mol k2 Ea = – R (ln ) k1 1 600 K – 1 – T2 1 500 K 1 T1
1
1 Figure 16.10 Informa=on sequence to determine the kine=c parameters of a reac=on. The dependence of number of possible collisions on the product of reactant concentra=ons. The Collision Theory • In a chemical
reac=on, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other. Figure 16.12 The eﬀect of temperature on the distribu=on of collision energies. Figure 16.13 Energy
level diagram for a reac=on. The forward reac=on is exothermic because the reactants have more energy than the products. Figure 16.14 The importance of molecular orienta=on to an eﬀec=ve collision. NO + NO3 2 NO2 A is the frequency factor A = pZ where Z is the collision frequency p is the orienta=on probability factor Figure 16.15 Nature of the transi=on state in the reac=on between CH3Br and OH
. CH3Br + OH
CH3OH + Br
transi7on state or ac7vated complex Figure 16.16 Reac=on energy diagram for the reac=on of CH3Br and OH
. Figure 16.17 Reac=on energy diagrams and possible transi=on states for two reac=ons. ...
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This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.
 Spring '08
 HOEGER
 Reaction

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